**Proof:** Let $\displaystyle F$ be all eventually zero rational sequences

$\displaystyle F=\{(r_1,r_2,\dots,r_n,0,0,\dots)|r_i \in \mathbb{Q},n\in \mathbb{N}\}.$

Clearly, $\displaystyle F \subseteq \ell^2$ and is countable. I claim $\displaystyle \text{cl}(F)=\ell^2$.

Let $\displaystyle y=(y_k)_{k \in \mathbb{N}} \in \ell^2$ and $\displaystyle \varepsilon>0.$ Pick $\displaystyle n_0 \in \mathbb{N}$ such that $\displaystyle \sum_{k=n_0+1}^{\infty}|y_k|^2 =M_{n_0} < \varepsilon^2.$ Since $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$, for each $\displaystyle y_k$ (for $\displaystyle 1\le k \le n_0$), pick $\displaystyle r_{0k} \in \mathbb{Q}$ such that

$\displaystyle |r_{0k}-y_k| < \sqrt{\frac{1}{n_0}\left(\varepsilon^2-M_{n_0}\right)} \Longrightarrow |r_{0k}-y_k|^2 < \frac{1}{n_0}\left(\varepsilon^2-M_{n_0}\right)$.

So we have

$\displaystyle \sum_{k=1}^{n_0}|r_{0k}-y_k|^2<\varepsilon^2-M_{n_0}$.

If we let $\displaystyle x_0 = (r_{01},r_{02},\dots,r_{0n_0},0,0,\dots) \in F$, then $\displaystyle d(x_0,y)<\varepsilon$. Therefore, $\displaystyle y$ is a limit point of $\displaystyle F$.